Joyce Smith

2021-12-11

Find the minimum value of $f\left(x\right)=2{x}^{3}+3$ in the interval [-2,2].
The absolute minimum value is 5 at x=1.
The absolute minimum value is 3 at x=0.
The absolute minimum value is -13 at x=-2.
The absolute minimum value is 19 at x=2.

Shannon Hodgkinson

Step 1
We first find the critical point by solving f'(x)=0
$f\left(x\right)=2{x}^{3}+3$
${f}^{\prime }\left(x\right)=6{x}^{2}$
$0=6{x}^{2}$
${x}^{2}=0$
x=0
Step 2
Then we check the values of f(x) at the critical point and the endpoints
$f\left(x\right)=2{x}^{3}+3$
At x=-2, f(-2)=-13
At x=0, f(0)=3
At x=2, f(2)=19
So the minimum value is -13.
Answer: The absolute minimum value is -13 at x=-2

Do you have a similar question?